Laminas on inclined planes and hinged laminas

Toppling on a slope

A lamina standing on a plane is on the point of toppling when the vertical line through its centre of mass G passes through the lowest point of contact (the lower corner). If the line falls outside the base, the weight has a moment about that corner and the lamina tips over.

For a rectangle of width w and height h resting on a plane at angle α: it is about to topple when tan α = (w/2)/(h/2) = w/h. In general tan α = (distance of G from the lower corner, measured along the plane) ÷ (height of G above the plane).

Slide or topple?

On a rough plane the lamina may slide first. Sliding starts when tan α > μ. Compare the angle for sliding with the angle for toppling; whichever is smaller happens first.

Hinged laminas

A lamina free to rotate about a horizontal axis through a point A is held in equilibrium by another force. Take moments about A so the unknown reaction at the axis does not appear. The weight acts at G; use the horizontal distance from A to G.

Worked example

A uniform rectangular block has cross-section 4 cm wide and 6 cm tall. It stands on a rough plane whose angle is slowly increased. Assuming it does not slide, find the angle at which it topples.

1. G is 2 cm from the lower edge along the plane and 3 cm above the plane.
2. About to topple when the vertical through G passes through the lower corner.
3. tan α = 2/3, so α = 33.7°.

Answer: 33.7°

Check yourself

1. A uniform square lamina stands on a rough inclined plane, in a vertical plane containing a line of greatest slope, with one edge on the plane. At what angle of the plane does it topple, assuming it does not slide?
   Show answer

   45 °. tan α = (s/2)/(s/2) = 1, so α = 45°.

2. A uniform rectangular lamina 3 cm wide and 8 cm tall stands with its 3 cm edge on a rough inclined plane, in a vertical plane containing a line of greatest slope. The angle of the plane is slowly increased and the lamina does not slide. Find the angle at which it topples, in degrees to 1 decimal place.
   Show answer

   20.6 °. tan α = 1.5/4, α = 20.6°.

3. The 4 cm by 6 cm block of the worked example is on a plane with μ = 0.5. As the angle increases, what happens first?
   • It topples, because 2/3 is less than 0.5
   • It topples and slides at the same angle
   • It never moves
   • It slides, because tan α reaches 0.5 before 2/3
   Show answer

   It slides, because tan α reaches 0.5 before 2/3. Sliding at tan α = 0.5 (26.6°) comes before toppling at tan α = 2/3 (33.7°).

4. A uniform right-angled triangular lamina has its 6 cm base on a rough inclined plane, in a vertical plane containing a line of greatest slope. Its 9 cm side is perpendicular to the plane at the upper end of the base. The lamina does not slide. Find the angle at which it topples about its lower corner, in degrees to 1 decimal place.
   Show answer

   53.1 °. G is 2 cm from the right angle, so 4 cm from the lower corner along the plane, and 3 cm above it. tan α = 4/3, α = 53.1°.

5. A uniform square lamina ABCD of side 2 m and weight 40 N is in a vertical plane, free to rotate about a horizontal axis through A, with AB horizontal. It is held there by a vertical force at B. Find this force.
   Show answer

   20 N. Moments about A: F × 2 = 40 × 1, so F = 20 N.

6. A lamina on an inclined plane is on the point of toppling. Where does the vertical line through its centre of mass pass?
   • Through the lowest point of contact with the plane
   • Through the highest point of contact
   • Through the middle of the base
   • Outside the base on the uphill side
   Show answer

   Through the lowest point of contact with the plane. That is the boundary between staying up and tipping over.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M2
• Pearson Edexcel IAL Mechanics M2 (WME02) question paper, June 2025: front-cover instructions on g and accuracy

Last updated 8 November 2026. Spotted a mistake? Tell us.